Intersections and Equations That Reduce to Quadratics — Integrated Mathematics

9.5 Intersections and Equations That Reduce to Quadratics

Many equations that are not quadratic at first sight become quadratic equations after a rearrangement or a change of variable. Two common cases are finding where two graphs meet and substituting a new variable for a repeated expression.

Intersections of functions

Note. If finding the intersection of two functions, then set the functions equal.

Example 9.5.1. Find the intersections of and .

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At and the functions meet at and .

Example 9.5.2. Find the intersections of and .

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using . So or .

Checking the -values:

: and .
: and .

The intersections are and .

Example 9.5.3. Find the intersections of and .

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(X: top , bottom , sides and ), so and .

Check : and .

Solving with substitutions

Example 9.5.4. Solve .

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For a substitution to work, the two expressions in brackets must be the same (here both are ), and the exponent on the first must be 2 times the exponent on the next (here and ). Let Then , and

So , which means

Example 9.5.5. Solve .

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Let : X: top , bottom , sides and (which gives ). So Back to :

Let’s say instead we had . Then which is real, since .

Example 9.5.6. Solve .

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Let : So Check : ? Since , this is .

Example 9.5.7. Solve .

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Let : So is a false solution: can never equal , and squaring to get hides this ().

Example 9.5.8. Solve .

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Rewritten: . Let : so Back to :

Exercises

Exercise 9.5.9. (a)

(b)

(c)

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(a) . Let . Recall .

By the formula,

Or: needs and . Splitting the middle term, which gives the factor .

Let’s restart to make it clean: , and

(b) , and (using ), so and .

(c) . Wait, ur stupid, duh:

Exercise 9.5.10. Find if and .

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Substituting : and , where has no real roots. So and .

Exercise 9.5.11. Find all ordered pairs with such that and .

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Let , so : So either , which means , or , which means , i.e. .

With : so or (check: ).

Substitute each value of back into .

If : so

If : so giving or .

Checks: : ✓, and : ✓. But has to hold too: ✓, while . So from these substitutions only and survive. The case makes and the roots of , i.e. , which gives and .

Final answer: YAY

Exercise 9.5.12. Find all such that , with .

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Both terms vanish (they become like and on each side) when and : with , (and , ).

Clearing denominators: the numerators become and , with

Another try: set the numerators’ sum to zero, where . Then Factoring the sum of the numerators instead gives , so , : again or .

Properly, factor out from both numerators: The bracket vanishes when so , i.e. , and besides , So the solutions are , , and .

Exercise 9.5.13. Let be the real roots of . Find .

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Let (zero at ). Then , and the left side is . where so